Download MODEL ANSWERS TO THE FIRST QUIZ 1. (18pts) (i) Give the

MODEL ANSWERS TO THE FIRST QUIZ
1. (18pts) (i) Give the definition of a m × n matrix.
A m × n matrix with entries in a field F is a function
A : I × J −→ F,
where I is the set of integers between 1 and m and J is the set of
integers from 1 to n.
(ii) Give the definition of row echelon form.
A matrix U is in echelon form if
• the first non-zero entry of every row, called a pivot, is 1.
• a row with a pivot is above a row without one.
• a pivot which comes above another pivot also occurs to the left
of that pivot.
(iii) Give the definition of the rank of a matrix.
The rank of a matrix A is the number of pivots in a matrix U in
row echelon form, which is also row equivalent to A.
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(iv) Give the definition of the left inverse of a matrix.
If A is a m × n matrix then a left inverse of A is a n × m matrix
B such that BA = In .
(v) Give the definition of linear combination.
A vector v ∈ V is a linear combination of v1 , v2 , . . . , vk ∈ V if
there are scalars r1 , r2 , . . . , rk such that v = r1 v1 + r2 v2 + . . . rk vk .
(vi) Give the definition of closed under addition.
Let V be a vector space. We say that a subset W ⊂ V is closed
under addition if whenever v and w belong to W then so does their
sum v + w.
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2. (15pts) (i) Let
V = { f ∈ Pd (R) | f 0 (0) = 0 },
be the set of real polynomials of degree at most d whose derivative at
zero is zero. Is V a subspace of Pd (R)?
True. It suffices to check that V is non-empty, closed under addition
and scalar multiplication. The zero polynomial has zero derivative at
zero, and so V is certainly non-empty. If f and g are two elements of
V then f 0 (0) = g 0 (0) = 0. But then
(f + g)0 (0) = f 0 (0) + g 0 (0) = 0,
using the standard rules for differentation. It follows that f + g ∈ V
and so V is closed under addition. Finally if f ∈ V and λ ∈ R then
f 0 (0) = 0. But then
(λf )0 (0) = λf 0 (0) = 0,
so that λf ∈ V . Hence V is closed under scalar multiplication.
(ii) Let F be a field and let
V = { A ∈ M2,2 (F ) | rk(A) ≤ 1 },
be the set of 2 × 2 matrices of rank at most one. Is V a subspace of
M2,2 (F )?
False. We have
1 0
0 0
1 0
+
=
.
0 0
0 1
0 1
Both matrices on the left have rank one but the matrix on the right has
rank two. Therefore V is not closed under addition and so it cannot
be a subspace.
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3. (10pts) Under what conditions on the real number c is the vector
(1, c, 1) ∈ R3 a linear combination of the vectors v1 = (−1, 3, 2) and
v2 = (2, −6, −5) ∈ R3 ?
Let A be the matrix whose columns are the vectors v1 and v2 . Then


−1 2
A =  3 −6 .
2 −5
Then v is a linear combination of v1 and v2 if and only if the equation
Ax = b has a solution. To determine this form the augmented matrix


−1 2 | 1
 3 −6 | c  ,
2 −5 | 1
and apply Gaussian elimination. Multiply the first row by −1.


1 −2 | −1
3 −6 | 1  .
2 −5 | c
Multiply the first row by −3 and −2 and add it the second and third
rows


1 −2 | −1
0 0 | c + 3  .
0 −1 |
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Finally swap the second and third rows and multiply the second row
by −1,


1 −2 | −1
0 1 | −3  .
0 0 | c+3
The system of equations represented by this matrix has the same solutions as the system represented by the original matrix. The last
equation can only be solved if c + 3 = 0 that is c = −3. In this case it
is easy to solve these equations using back substitution. (As it happens
r2 = −3 and r1 = −7 so that (1, −3, 1) = −7(−1, 3, 2) − 3(2, −6, −5)).
The span of v1 and v2 is a plane. Vectors of the form (1, c, 1) form
a line. The line and the plane intersect in a point (namely (1, −3, 1)),
since the line is neither contained in the plane nor parallel to it.
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4. (25pts) Let A be the real 3 × 4 matrix


1 −2 3 −1
−2 4 −6 2  .
5 −10 16 4
(i) Express the matrix A as a product A = P LU of a permutation
matrix P , a lower triangular matrix L and a matrix U in echelon form.
We apply a modified version of Gaussian elimination. We recognise
that the second row is a multiple of the first. So we first swap the
second and third rows. This is the only permutation of the rows we
need, so we have


1 0 0
P = P2,3 = 0 0 1 .
0 1 0
After swapping the second and third rows of A we have


1 −2 3 −1
 5 −10 16 4  .
−2 4 −6 2
Multiply the first row of A by −5 and 2 and add it to the second and
third row


1 2 3 −1
U = 0 0 1 −1 .
0 0 0 0
As this completes the elimination, U is the indicated matrix. Finally,


1 0 0
L =  5 1 0 .
−2 0 1
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(ii) Find the kernel of A, that is the set of solutions to the homogeneous equation Ax = 0.
This is the same as the kernel of U . We solve the system U x = 0 by
back substitution. If the variables as (a, b, c, d) then b and d are free
variables. The second equation determines c in terms of d, c − d = 0,
that is c = d. The first equation then reads a + 2b + 3d − d = 0, so
that a = −2b − 2d. Therefore
(−2b − 2d, b, d, d) = b(−2, 1, 0, 0) + d(−2, 0, 1, 1),
is the general solution to the homogeneous equation.
(iii) Given that v = (1, 1, 1, 1) is a solution of the equation Ax = b,
where b = (1, −2, 15), what is the general solution?
(1, 1, 1, 1) + b(−2, 1, 0, 0) + d(−2, 0, 1, 1).
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5. (30pts) For each statement below, say whether the statement is
true or false. If it is false, give a counterexample and if it is true then
explain why it is true. Let A be m × n matrix with entries in a field.
(i) If A m < n then the equation Ax = b always has a solution.
False. Let A = (0, 0), b = (1). Then m = 1 < 2 = n and it is clear
that the equation Ax = b has no solutions.
(ii) Let φ : Fn −→ F m be the function φ(v) = Av. If φ is surjective
then the equation Ax = b always has a solution.
True. Pick b ∈ F m . As φ is surjective there is a vector v ∈ F n such
that φ(v) = b. Then Av = φ(v) = b, so that v ∈ F n is a solution of the
equation Ax = b.
(iii) If B is an invertible n × n matrix then the nullspace of A and
the nullspace of AB are the same.
False. Let A = (1, 0) and let
0 1
B=
1 0
Then B 2 = I2 so that B is invertible. The kernel of A is
{ v = (x, y) ∈ R2 | x = 0 },
and the kernel of AB = (0, 1) is
{ v = (x, y) ∈ R2 | y = 0 },
which are clearly not equal.
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(iv) If B is an invertible m × m matrix then the nullspace of A and
the nullspace of BA are the same.
True. Suppose that v ∈ Ker A. Then
(BA)v = B(Av) = B0 = 0.
Thus v ∈ Ker BA and so Ker A ⊂ Ker BA. Let C be inverse of B.
By what we already just proved Ker BA ⊂ Ker C(BA). But C(BA) =
(CB)A = A. So Ker BA ⊂ Ker A. Since we have an inclusion either
way, we have Ker A = Ker BA.
(v) A matrix A can have at most one inverse.
True. Let B and C be two inverses of A. We compute B(AC). Since
C is an inverse of A, B(AC) = BIn = B. On the other hand, as matrix
multiplication is associative we have B(AC) = (BA)C. As B is the
inverse of A, (BA)C = In C = C.
(vi) Suppose that the entries of A are rational numbers. If the equation Ax = b has a real solution v ∈ Rn then it has a rational solution
w ∈ Qn .
True. Let C = (A|b) be the augmented matrix. We apply Gaussian
elimination. At the end we get (U |c), row equivalent to A, where U is
in echelon form. The equation U x = c has the same real and rational
solutions as the equation Ax = b, and U is a matrix with rational
entries. Since the equation U x = c has a real solution, it follows that
every row of zeroes of U is also a row of zeroes of the augmented matrix
(U |c).
But then we could solve the equation U x = c using back substitution.
If we pick rational numbers for the free variables then we get rational
solutions.
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